Inverse of a function

**write2rvr** · March 16th 2009, 05:48 AM

f(x)=[x^2+5]/[x-3] where x not equal to 3

**e^(i*pi)** · March 16th 2009, 06:02 AM

Originally Posted by write2rvr

f(x)=[x^2+5]/[x-3] where x not equal to 3

set $f(x) = y$ and make x the subject:

$yx - 3y = x^2 + 5$

$x^2 - yx = -(5+3y)$

$(x+\frac{y}{2})^2 - \frac{y^2}{4} = -5 - 3y$ <-- from completing the square

$(x+\frac{y}{2}) = ±\sqrt{\frac{y^2}{4} - 3y - 5}$

$x = - \frac{y}{2} ±\sqrt{\frac{y^2}{4} - 3y - 5}$

Now set any y to x and set x to $f^{-1}(x)$

$f^{-1}(x) = - \frac{x}{2} ±\sqrt{\frac{x^2}{4} - 3x - 5}$

**stapel** · March 16th 2009, 09:58 AM

Originally Posted by write2rvr

f(x)=[x^2+5]/[x-3] where x not equal to 3

When you get to:

. . . . .0 = x^2 - xy + 3y + 5

...you can always plug this into the Quadratic Formula, using a = 1, b = -y, and c = 3y + 5.

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